Index two.

[ForMyThunder]

Is it true that if a finite group G contains a subgroup of index 2, then there is an element of G with order 2?

[ForMyThunder]

I'm thinking that if H is a subgroup of index 2, then H is normal in G. Then for all x in G, xH=Hx, so (xH)(Hx)=H and xHx=H. So x=x^-1 and x has order 2. Is this correct?

[Newtime]

The statement is true. If G has finite order and a subgroup of index two, G must have even order by Lagrange's theorem. Then, you can easily prove there must exist an element which is its own inverse. In fact, this is a standard exercise in any text.

Regarding your proof: Is multiplication of a left coset by a right coset well defined? And why do you end up with H on the right side? Doing this assumes xx=e.

[ForMyThunder]

You're right.

But you have to have x^2 in H for all x not in H. (xH)(xH)=H because otherwise, it would be an identity: (xH)(xH)=(xH) and cancellation gives that x is in H.

I don't see any way to prove this.